Definition of Besov space $B^\alpha_{2,\theta}(-\pi,\pi)$ and embeddings

Question

I consider Besov space $B^\alpha_{2,\theta}(-\pi,\pi)$ defined as follows for $2\pi$-periodic functions, \begin{align} B^\alpha_{2,\theta}(-\pi,\pi)=\left\{f~\Bigg|~\sum_{j=0}^{\infty}\left(\sum_{2^j\leq |n| < 2^{j+1}} |n|^{2\alpha} |c_n(f)|^2 \right)^{\theta/2}<\infty\right\} \end{align} for $\alpha\geq0$ and $1\leq \theta < \infty$. Now I have tried to prove that $B^0_{2,\theta}(-\pi,\pi) \subset L^2(-\pi,\pi)$ for $1\leq \theta \leq 2$, and $L^2(-\pi,\pi) \subset B^0_{2,\theta}(-\pi,\pi)$ for $2\leq \theta <\infty$. At first, I thought I had completed the proof as follows. Since there is $j_0 \in \mathbb{N}$ such that $\left(\sum_{2^j \leq |n| < 2^{j+1}} |c_n(f)|^2\right)^{1/2} \ll 1$ for $j_0 \leq j$, we have \begin{align} \sum_{j_0 \leq j}\left(\sum_{2^j \leq |n| < 2^{j+1}} |c_n(f)|^2 \right)^{\theta_1/2} \leq \sum_{j_0 \leq j}\left(\sum_{2^j \leq |n| < 2^{j+1}} |c_n(f)|^2 \right)^{\theta_2/2} < \infty\,, \end{align} for any $1 \leq \theta_2 \leq \theta_1$. Subsequently, we have \begin{align} \sum_{j=0}^\infty\left(\sum_{2^j \leq |n| < 2^{j+1}} |c_n(f)|^2 \right)^{2/2} = \sum_{n=-\infty}^\infty |c_n(f)|^2 = \frac{1}{2\pi}\int_{-\pi}^\pi |f(x)|^2 dx < \infty\,, \end{align} for $\theta=2$. However, I realized that the definition of $B^\alpha_{2,\theta}(-\pi,\pi)$ does not cover the case $n=0$. But I think both $B^0_{2,2}(-\pi,\pi)$ and $L^2(-\pi,\pi)$ must be the same space. If so, is the definition wrong or am I just not sure how to prove it?

Answer 1

After long days of worries, I am convinced that there is a problem unless $f\in L^2(-\pi,\pi)$ already since that definition doesn't include the origin. Therefore, we must slightly modify the range of $\vert n \vert$, for instance, $2^j-1$ to $2^{j+1}-2$ or just give information of $c_0(f)$ explicitly.